package com.alexd.leetcode.contest.week118;

import java.util.*;

public class Solution1 {

/**
 * 	
 * 给定两个非负整数 x 和 y，如果某一整数等于 x^i + y^j，其中整数 i >= 0 且 j >= 0，那么我们认为该整数是一个强整数。

返回值小于或等于 bound 的所有强整数组成的列表。

你可以按任何顺序返回答案。在你的回答中，每个值最多出现一次。

 

示例 1：

输入：x = 2, y = 3, bound = 10
输出：[2,3,4,5,7,9,10]
解释： 
2 = 2^0 + 3^0
3 = 2^1 + 3^0
4 = 2^0 + 3^1
5 = 2^1 + 3^1
7 = 2^2 + 3^1
9 = 2^3 + 3^0
10 = 2^0 + 3^2
示例 2：

输入：x = 3, y = 5, bound = 15
输出：[2,4,6,8,10,14]
 

提示：

1 <= x <= 100
1 <= y <= 100
0 <= bound <= 10^6
 */
	
	public List<Integer> powerfulIntegers(int x, int y, int bound) {
		Map<Integer, Integer> yMap = new HashMap<Integer, Integer>();
		Map<Integer, Integer> xMap = new HashMap<Integer, Integer>();
		
		int yNum = 1;
		for (int yLoop = 0; yNum < bound; yLoop++) {
			yMap.put(yLoop, yNum);
			yNum = yNum *y;
			if (y == 1) {
				break;
			}
		}
		
		int xNum = 1;
		for (int xLoop = 0; xNum < bound; xLoop++) {
			xMap.put(xLoop, xNum);
			xNum = xNum *x;
			if (x == 1) {
				break;
			}
		}
		
		Set<Integer> res = new HashSet<Integer>();
		for (int i = 0; i < xMap.size() && xMap.get(i) < bound ; i++) {
			int xValue = xMap.get(i);
			for (int j = 0; j < yMap.size() && yMap.get(j) + xValue <= bound ; j++) {
				int resValue = yMap.get(j) + xValue;
				System.out.println("resValue=" + resValue + "j=" + j);
				res.add(resValue);
			}
		}
		
		return new ArrayList<Integer>(res);
        
    }	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		//System.out.println(new Solution1().powerfulIntegers(3, 5, 15));
		
		System.out.println(new Solution1().powerfulIntegers(1, 2, 100));
	}

}
